10th Class Mathematics Triangles Question Bank Case Based (MCQs) - Triangles

  • question_answer
    In a \[\Delta ABC,\] \[BD\bot AC\] such that \[B{{D}^{2}}=DC.\] AD, then :

    A) \[\angle A=90{}^\circ \]

    B) \[\angle B=90{}^\circ \]

    C) \[\angle C=90{}^\circ \]

    D) ABC is not a right angled triangle

    Correct Answer: B

    Solution :

    It is given that in triangle ABC, \[BD\bot AC\] such that
    Therefore, applying Pythagoras theorem in        triangles ABD and BCD, we get                 
    \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\]                    ...(1)    
    \[B{{C}^{2}}=B{{D}^{2}}+D{{C}^{2}}\]                    ...(2)   
    Adding equations (1) and (2), we get,
    \[=2(DC.AD)+A{{D}^{2}}+D{{C}^{2}}\] \[[B{{D}^{2}}=DC.AD]\]
    Therefore, \[A{{B}^{2}}+B{{C}^{2}}={{(AD+DC)}^{2}}=A{{C}^{2}}\]
    This means that by converse of Pythagoras theorem, ABC is a right triangle with AC os the hypotenuse and angle opposite to AC is \[90{}^\circ \]. Therefore, \[\angle B=90{}^\circ \].
    So, option [b] is correct.

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